Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
int n, m, q, cnt = 0;
int size[MAXN], son[MAXN], fa[MAXN], top[MAXN], dep[MAXN], id[MAXN], pre[MAXN];
vector<int> city[MAXN];
int esize = 0;
int head[MAXN];
struct EDGE {
  int u, v;
  int next;
} edge[MAXN << 1];
struct Vector {
  int len, Q[11];
  Vector() {
    len = 0;
    memset(Q, 0, sizeof(Q));
  }
  inline void push_back(int x) { Q[len++] = x; }
  inline int size() { return len; }
  inline void clear() {
    len = 0;
    memset(Q, 0, sizeof(Q));
  }
} ve[MAXN << 2];
inline int read() {
  int X = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    X = (X << 1) + (X << 3) + (ch ^ '0');
    ch = getchar();
  }
  return X * f;
}
inline void addedge(int u, int v) {
  edge[esize].u = u;
  edge[esize].v = v;
  edge[esize].next = head[u];
  head[u] = esize++;
}
void dfs1(int u) {
  size[u] = 1;
  for (int i = head[u]; ~i; i = edge[i].next) {
    if (edge[i].v == fa[u]) continue;
    fa[edge[i].v] = u;
    dep[edge[i].v] = dep[u] + 1;
    dfs1(edge[i].v);
    size[u] += size[edge[i].v];
    if (size[edge[i].v] > size[son[u]]) son[u] = edge[i].v;
  }
}
void dfs2(int u, int topf) {
  top[u] = topf;
  id[u] = ++cnt;
  pre[id[u]] = u;
  if (!son[u]) return;
  dfs2(son[u], topf);
  for (int i = head[u]; ~i; i = edge[i].next) {
    if (edge[i].v == fa[u]) continue;
    if (edge[i].v == son[u]) continue;
    dfs2(edge[i].v, edge[i].v);
  }
}
Vector merge(Vector B, Vector C) {
  Vector A;
  int i = 0, j = 0;
  while (A.size() < 10 && i < B.size() && j < C.size()) {
    if (B.Q[i] < C.Q[j])
      A.push_back(B.Q[i++]);
    else
      A.push_back(C.Q[j++]);
  }
  while (A.size() < 10 && i < B.size()) A.push_back(B.Q[i++]);
  while (A.size() < 10 && j < C.size()) A.push_back(C.Q[j++]);
  return A;
}
void buildtree(int l, int r, int rt) {
  if (l == r) {
    for (vector<int>::iterator it = city[pre[l]].begin();
         it != city[pre[l]].end(); ++it) {
      ve[rt].push_back(*it);
      if (ve[rt].size() == 10) break;
    }
    return;
  }
  int mid = (l + r) >> 1;
  buildtree(l, mid, rt << 1);
  buildtree(mid + 1, r, rt << 1 | 1);
  ve[rt] = merge(ve[rt << 1], ve[rt << 1 | 1]);
}
Vector query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) return ve[rt];
  int mid = (l + r) >> 1;
  Vector ans;
  if (L <= mid) ans = merge(ans, query(L, R, l, mid, rt << 1));
  if (mid + 1 <= R) ans = merge(ans, query(L, R, mid + 1, r, rt << 1 | 1));
  return ans;
}
Vector qRange(int u, int v) {
  Vector ans;
  while (top[u] != top[v]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    ans = merge(ans, query(id[top[u]], id[u], 1, n, 1));
    u = fa[top[u]];
  }
  if (dep[u] > dep[v]) swap(u, v);
  ans = merge(ans, query(id[u], id[v], 1, n, 1));
  return ans;
}
void readdata() {
  memset(head, -1, sizeof(head));
  n = read();
  m = read();
  q = read();
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read();
    addedge(u, v);
    addedge(v, u);
  }
  for (int i = 1; i <= m; ++i) {
    int x = read();
    city[x].push_back(i);
  }
}
void work() {
  dfs1(1);
  dfs2(1, 1);
  buildtree(1, n, 1);
  while (q--) {
    int u = read(), v = read(), a = read();
    Vector ans = qRange(u, v);
    int len = min(a, ans.size());
    printf("%d", len);
    for (int i = 0; i < len; ++i) printf(" %d", ans.Q[i]);
    printf("\n");
  }
}
int main() {
  readdata();
  work();
  return 0;
}